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aleksley [76]
4 years ago
13

Two pressure gauges measure a pressure drop of 16.3 psi (lb/in.2) at the entrance and exit of an old buried pipeline. The origin

al drawings have been lost. If the 6-in. galvanized iron pipe carries water at 68°F with a flow rate of 1.64 cfs (ft3/s), determine the length of the horizontal underground line ignoring minor losses.
Engineering
1 answer:
Sholpan [36]4 years ago
7 0

Answer: 866.25ft

Explanation:

Pressure drop= 16.3 psi

1 feet of water column= 0.4335psi

x= 16.3psi

Therefore, head loss of water= 16.3/0.4335=37.6ft

In inches, head loss of water= 37.6 × 12=451.2 inches

Q= 1.64ft^2/s

A= pi/4 D^2

Where D= 6 inches, 0.5ft

A= 3.142/4 × (0.5)^2

A= 0.1962ft^2

V= Q/A= 1.64/ 0.1962

V= 8.35ft/s, 100.28 in/s

REGNOLD'S NO

Re= SvD/ μ=

Where, D= 6 inches, V= 100.28 inches/s, S=62.4lb/ft^3= 0.03611lb/inches^3

At 68F, Dynamic viscosity μ= 1.0016× 10^-3NS/m^2

1psi= 6895N/m^3

μ= 1.0016× 10^-3NS/m^2/6895N/m^3

μ= 1.452× 10^-7lbs/ins^2

Re= 0.03611× 100. 28×6/1.42×10^-7

Re= 1.49×10^8

e/D----- Relative roughness

e= 0.0005ft

e/D= 0.001

Therefore, for Re= 1.49×10^8 and e/D= 0.001

F=0.02

Head loss is given by Dancy-Weisbach formula

hL= fLV^2/2gD

Wherr g= 9.81m/s^2= 386.12inches/s^2

451.2= 0.02×L×(100.28^2)/2 × 386.12inches/s^2× 6

L= 10395inches

L= 10395/12= 866.25ft

Therefore, the length of pipe is 866.25ft.

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Answer:

Explanation:

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Find the actual workdone by the compressor

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Find the temperature at exit of the turbine

T_4=\frac{1800}{8^{\frac{1.667-1}{1.667} }} \\\\=787.3k

Find the actual workdone by the turbine

1 \times 5.19 (1800-783.3)\\=5276.6kJ/kg

Find the temperature of the regeneration

\epsilon = \frac{T_5-T_2}{T_4-T_2} \\\\0.75=\frac{T_5-689.3}{783.3-689.3} \\\\T_5=759.8k

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Find the thermal efficiency

n_t_h=\frac{W_t-W_c}{Q_i_n} \\\\=\frac{5276.6-2020.4}{5388.2} \\\\n_t_h=60.4

60.4%

Find the mass flow rate

m=\frac{W_net}{P} \\\\\frac{60 \times 10^3}{5276.6-2020.4} \\\\=18.42

Find the actual workdone by the compressor

\frac{W_c}{m} =\frac{(\frac{W}{m} )}{n_c} \\\\=\frac{2020.4}{0.8} \\\\=2525.5kg

Find the actual workdone by the turbine

\frac{W_t}{m} =n_t(\frac{W}{m} )\\\\=0.8 \times5.19(1800-783.3)\\\\=4221.2kJ/kg

Find the temperature of the compressor exit

\frac{W_t}{m} =c_p(T_2_a-T_1)\\2525.5=5.18(T_2_a-300)\\T_2_a=787.5k

Find the temperature at the turbine exit

4221.2=5.18(1800-T_4_a)\\\\T_4_a=985k

Find the temperature of regeneration

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Write a static method called quadrant that takes as parameters a pair of real numbers representing an (x, y) point and that retu
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Full Question

Write a static method called quadrant that takes as parameters a pair of real numbers representing an (x, y) point and that returns the quadrant number for that point.

Recall that quadrants are numbered as integers from 1 to 4 with the upper-right quadrant numbered 1 and the subsequent quadrants numbered in a counter-clockwise fashion. Notice that the quadrant is determined by whether the x and y coordinates are positive or negative numbers. If a point falls on the x-axis or the y-axis, then the method should return 0.

Answer

public int quadrant(double x, double y) // Line 1

{

if(x > 0 && y > 0){ //Line 2

return 1;

}

if(x < 0 && y > 0){ // Line 3

return 2;

}

if(x < 0 && y < 0) {// Line 4

return 3;

}

if(x > 0 && y < 0) {// Line 5

return 4;

}

return 0;

}

Explanation:

Line 1 defines the static method.

The method is named quadrant and its of type integer.

Along with the method definition, two variables x and y of type double are also declared

Line 2 checks if x and y are greater than 0.

If yes then the program returns 1

Line 3 checks if x is less than 0 and y is greater than 0

If yes then the program returns 2

Line 4 checks if x and y are less than 0

If yes then the program returns 3

Line 5 checks is x is greater than 0 and y is less than 0

If yes then the program returns 4

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Quadrilateral ABCD is a rectangle.<br> If m ZADB = 7k + 60 and mZCDB = -5k + 40, find mZCBD.
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In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of the line has resis
Rufina [12.5K]

Question:

In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of the line has resistance R.

For a power line that supplies power to 10 000 households, we can conclude that

a) IV < I²R

b) I²R = 0

c) IV = I²R

d) IV > I²R

e) I = V/R

Answer:

d) IV > I²R

Explanation:

In a typical transmission line, the current I is very small and the voltage V is very high as to minimize the I²R losses in the transmission line.

The power delivered to households is given by

P = IV

The losses in the transmission line are given by

Ploss = I²R

Therefore, the relation IV > I²R  holds true, the power delivered to the consumers is always greater than the power lost in the transmission line.

Moreover, losses cannot be more than the power delivered. Losses cannot be zero since the transmission line has some resistance. The power delivered to the consumers is always greater than the power lost in the transmission.

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3 years ago
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